A comment on A000096

[Andrew Plewe]

Last week I was playing with triangle numbers and found the following interesting algorithm.

1.) Input your number.  In this case, I'll use f = 33.
2.) Find the nth triangle number t closest to f. In this case n = 7 and t = 28.
3.) Compute the remainder r = f - t. In this case 33 - 28 = 5.
4.) Plug r and n into the following quadratic equation:

 -3 +/- sqrt(9 + 8 * (n - r))
----------------------------- = q
              2

The values for n - r which produce perfect squares are 2, 5, 9, 14, 20, etc., or A000096.

By plugging in and solving, we get two results; q = (-3 +/- 5) / 2, or 1 and -4.

Take the positive result, 1, and subtract this from n (n - q = s), which gives 7 - 1 = 6.  If the result is even, then a GCD of (s,
n) will produce a divisor of n.  If the result is odd, s will divide n.  In our example, n is even, and GCD(6, 33) = 3.


I don't know if this is anything new or not, but it does provide an interesting comment for A000096 and a very fast method for
factoring some numbers (I've been playing with the derivation of this to see if I can come up with a more general solution).  I
wanted to run this by the list before submitting it just in case I've made a mistake of some sort.  If you're interested, I can send
a more involved derivation of the above quadratic equation.  Also I'm a bit intrigued by the negative result that comes out of the
equation, any ideas on what it means?  In the above example, subtracting -4 from 7 gives us 11, which evenly divides 33 but I'm not
sure if this is true for all valid integer solutions of the quadratic equation.  Thanks!

	-Andrew Plewe-